Question:
ABC is an equilateral triangle of side 10 m and D is mid point of BC. charges of 100, 100, 75 mu c are placed at B, C, d respectively. Find force on 1 mu c charge placed at A..
Answer:
ABC is an equilateral triangle of side 10m.
D is the midpoint of BC.
Charge on B = +100µC ; Charge on C = -100µC ; Charge on D = +75µC
Consider, put A at the top, D directly below.
B and C have equal charge;
lets find the magnitude of each of those forces.
F = kQq / d² = 9e9 N·m²/C² * 1e-6C * 100e-6C / (10m)² = 9 mN
The vertical components of these two forces (A-B and A-C) cancel; the horizontal forces add:
Fh = 2 * 9mN * cos60º = 9 mN to the right.
For D, the distance A-D is d = 10m * √3/2, so d² = 75 m², and
F = 9e9 N·m²/C² * 1e-6C * 75e-6C / 75m² = 9 mN up
So, the net force on A is
F = √(9² + 9²) mN = 9√2 mN = 9√2e-3 N
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